Need help with this problem A vapor power plant operating at steady state uses a
ID: 1862504 • Letter: N
Question
Need help with this problem
A vapor power plant operating at steady state uses a Rankine power cycle with water as the working substance at a mass flow rate of 80 kg/s. Superheated steam exits the boiler at 480 degree C, 8 MPa. Heat transfer and frictional effects in the line connecting the boiler to the turbine reduce the temperature and pressure to 440 degree C, 7.6 MPa. The turbine outlet pressure is 10 kPa. Liquid leaves the condenser at 36 degree C, 8 kPa. Cooling water to the condenser enters at 15 degree C, 1 bar and exits at 35 degree C, 1 bar. The pressure at the pump outlet is 8.6 MPa. The turbine and pump isentropic efficiencies are 0.88 and 0.83, respectively. Determine the net power output of the plant, the thermal efficiency, the rate of heat transfer from the line connecting the boiler to the turbine, and the mass flow rate of the cooling water to the condenser. Sketch the cycle on a T-s diagram. Neglect heat transfer from the turbine, condenser, and pump to their surroundings and changes in kinetic and potential energy.Explanation / Answer
Given:
m = 80 kg/s
P1 = 8 MPa
T1 = 480 deg C
P2 = 7.6 MPa
T2 = 440 deg C
P3 = 10 kPa
P4 = 8 kPa
T4 = 36 deg C
P5 = 8.6 MPa
T6 = 15 deg C
T7 = 35 deg C
Turb eff = 0.88
Pump eff = 0.83
From steam table,
For P1,T1 we get h1 = 3350 kJ/kg
For P2,T2 we get h2 = 3250 kJ/kg, s2 = 6.55 kJ/kg-K
For P3, s3s (= s2), we get h3s = 2070 kJ/kg
For P4, T4 we get h4 = 151 kJ/kg, s4 = 0.519 kJ/kg-K
For P5, s5s (= s4), we get h5s = 160 kJ/kg
Turb eff = (h2 - h3) / (h2 - h3s)
0.88 = (3250 - h3) / (3250 - 2070)
h3 = 2211.6 kJ/kg
Pump eff = (h5s - h4) / (h5 - h4)
0.83 = (160 - 151) / (h5 - 151)
h5 = 161.8 kJ/kg
Turbine output = h2 - h3 = 3250 - 2211.6 = 1038.4 kJ/kg
Pump input = h5 - h4 = 161.8 - 151 = 10.8 kJ/kg
Net output = 1038.4 - 10.8 = 1027.6 kJ/kg
Power output = m*1027.6 = 80*1027.6 = 82208 kW = 82.2 MW
Heat input = m*(h1 - h5) = 80*(3350 - 161.8) = 255056 kW = 255.06 MW
Thermal efficiency = Power output / Heat input = 82.2 / 255.06 = 0.322 or 32.2 %
Heat loss between boiler and turbine = m*(h1 - h2) = 80*(3350 - 3250) = 8000 kW = 8 MW
Cp of cooling water = 4.18 kJ/kg-C
Energy baance in heat exchanger:
m_cool*4.18*(T7 - T6) = m*(h3 - h4)
m_cool*4.18*(35 - 15) = 80*(2211.6 - 151)
m_cool = 1971.9 kg/s